# Electric Field due to Point Charges

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1. Nov 3, 2014

### Caiti

1. The problem statement, all variables and given/known data

The following is on a practice exam I have been completing. In advance, it is part b I am struggling with.

Two point charges Q1 = +5*10^(-6)C and -2*10^(-6)C are 50cm apart.
a) Where along a line that passes through the two charges is the electric potential zero (apart
from at r=infinity)?

b) Determine the magnitude of the electric field at that point. Explain why the electric field is non-zero.

2. Relevant equations
E=kQ/r^2

V=kQ/r

E=dV/dr

3. The attempt at a solution
I got part A immediately using V=kQ/r and superposition of electric potential, finding it to be 0.36m (which matched the solution). It is part b that I am struggling with. I have as follows:

E=kQ/r^2
Using superposition of electric fields, this gives
E = k(5/(0.36^2) + -2/(0.5-0.36)^2), where k=1/(4*pi*E_0). This gave me an electric field of -570367N/C; however, the solution said it was 1.2*10^6 N/C. I believe the issue was in my implementation of the -2C in the above equation; however, am not entirely sure why this is incorrect or what to do to fix it.

As for the explanation, my belief is that because E is changing in potential with respective to distance, as long as the potential is not set at a constant 0 - that is, the potential is changing as distance changes - the electric field could be non-zero. I think I'm okay with this, but please do not hesitate to correct me if I'm wrong!

I'd really appreciate some help in figuring this out - or even just in where I went wrong. Please let me know if I've forgotten anything or my working isn't clear.

Thank you so much

2. Nov 3, 2014

### robotpie3000

You are right there was something wrong with the use of the -2 x 10-6C. The equation should be E=k[((5x10-6)/0.362)+((2x10-6)/0.142)], because we are finding the magnitude of the electric field, so the plus and minus signs doesn't matter.

Last edited: Nov 3, 2014
3. Nov 3, 2014

### Caiti

So does that mean that if we were finding the electric field, not magnitude, I would be correct? Or would I still use E=k[((5x10-6)/0.362)+((2x10-6)/0.142)] , then choose direction afterwards?